ANSWERS & MARK SCHEMES

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Raymond Doyle
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1 QUESTIONSHEET 1 (a) gene mutations alter the base sequence on DNA thus altering the genetic code; chromosomal mutations alter the chromosome structure thus altering the sequences of genes on the chromosome; chromosome mutations may also alter the chromosome number of the individual; 3 (b) substitution/deletion/addition/inversion/translocation;;(any two) sickle cell anaemia/albinoism/melanism/any other valid example;;(any two) 4 (c) when an individual receives three copies of chromosome 21 instead of two; reference to non disjunction/translocation; reference to learning difficulties/thick set bodies/thick necks/infertility/any other correct symptom; max 2 (d) one gene mutation will produce a new character which will thus be a discontinuous variant; thousands of similar gene mutations will form a wide range of slightly different characters which will give continuous variation; 2 TOTAL 11 QUESTIONSHEET 2 increase in chromosome number; either of individual chromosomes or of complete sets; 2 allopolyploidy; 1 (iii) the genomes of one set of (the species ) chromosomes; 1 (iv) where the chromosomes fail to separate during anaphase; due to failure of the spindle to contract; thus new nuclear membrane forms round both lots of chromosomes together; max 2 (b) the individuals arising from polyploidy can all interbreed to form fertile offspring (and thus are a species); they cannot breed back to the parental stock to produce fertile offspring and are thus separate species to the parents; due to the impossibility of chromosomes to pair/form bivalents in meiosis; 3 (c) (i) they possess the genes of both parental species including the genes that confer the best/strongest characters; thus have a double dose of valuable genes; giving them greater survival potential; 3 increased yield; easier to thresh/extract grain/ better flour/higher disease resistance/quicker maturing/better nutritive value; 2 TOTAL 14 Page 1

2 QUESTIONSHEET 3 (a) mosquitoes were exposed to DDT to eradicate them as malarial vectors; alleles mutated in a few mosquitoes and gave them DDT resistance; these resistant forms survived and reproduced to form more resistant mosquitoes; because non-resistant forms had been wiped out there was no competition; and resistant forms could flourish/less selection pressure on resistant forms; max 4 (b) over use of antibiotics by medical profession/in animal foodstuffs exposed many bacteria to antibiotics; mutant alleles appeared which gave certain strains of bacteria antibiotic resistance; the mutant alleles were in the plasmid DNA; when bacteria die the plasmids are released into the substrate; and may become incorporated into other species of bacteria (thus giving them antibiotic resistance); max 4 (c) due to the formation of a restitution nucleus/equivalent, that doubles the chromosome number; thus it can now form bivalents in meiosis and so produce gametes; 2 TOTAL 10 QUESTIONSHEET 4 (a) radiation that is naturally present in the environment; comes from cosmic rays hitting Earth/ from radioactive elements (such as uranium/thorium/radon) in the Earth s rocks /from natural radioactive carbon and potassium isotopes in biological matter; 2 (b) alpha = helium nuclei; beta = electrons/positrons; gamma = electromagnetic energy; beta is least likely since its particles are light and have little penetrating power; 4 (c) mustard gas; 5-bromouracil/dioxin/any other valid example; 2 (d) (i) there is a linear relationship between the dose of radiation over a total life span and the amount of mutation /the higher the dose the more mutation and time between doses does not limit the mutation; 1 X = 12.5 x 2; = 5%; 2 5 (e) (i) Any three of: substitution/addition/deletion/inversion/translocation; 1 Any three of: deletion/addition/inversion/translocation; 1 TOTAL 13 Page 2

3 QUESTIONSHEET 5 cells (in culture) treated with colchicine/drug; which holds chromosomes in metaphase of mitosis/in their most visible state; cells smeared on slide, (fixed) and stained to show chromosomes; chromosomes in many nuclei photographed through microscope; chromosomes cut out singly and matched into pairs; according to length/shape/position of centromere/staining pattern; max 4 amniotic fluid sample taken from pregnant mother; this contains fetal cells/cells shed from the baby; these can be gently centrifuged to concentrate them; then placed in tissue culture; allowed to grow and divide (mitotically) for several days; max 3 yes; 46 plus 1 extra chromosome/22 pairs plus a group of 3 identical chromosomes; 2 trisomy 21/Downs syndrome; mutation; ref to non disjunction; chromosomes 21 failed to separate in anaphase and both went to same pole/egg nucleus; thus when egg was fertilised the zygote contained three of chromosome 21; max 4 (iii) female; if it was male a different shaped Y chromosome would be visible; 2 TOTAL 15 Page 3

4 QUESTIONSHEET 6 DNA triplet would become CAT; this would form GUA by transcription to the mrna; GUA codes for valine; thus sixth amino acid in chain/penultimate amino acid is changed from glutamic acid to valine; max 3 will affect/alter the cross-bonding in the globin chain/polypeptide; which will alter the 3-D shape/conformation/tertiary structure of the molecule; 2 (iii) substitution; 1 (b) malarial parasite develops inside red cells of humans; cannot survive on haemoglobin S as substrate/cannot survive in reduced potassium ion environments; thus sickle cell sufferers are resistant to malarial infection but usually die from sickle cell anaemia at an early age; sickle cell trait heterozygotes are resistant to malarial infection and do not die from sickle cell anaemia; thus reproduce normally raising incidence of mutant gene in the population of the malarial zone; (probably) have a greater reproductive capacity than malarial sufferers (within the population); max 4 TOTAL 10 QUESTIONSHEET 7 (three) nucleotides could be omitted during replication of DNA (in meiosis/gamete formation); thus mrna does not include the omitted nucleotides (during transcription); thus an amino acid (actually phenylalanine) will be omitted from the polypeptide chain; during translation; thus CFTR protein will not work/work properly/be ineffective; 4 substitution; insertion/addition; translocation; max 2 (use suitable symbols, eg C for normal allele, c for cystic fibrosis allele) P Cc x Cc ; gametes C c C c ; F 1 CC Cc Cc cc ; normal carriers sufferer ; probabability 1 in 4 ; 5 CF males are sterile so their genes can be disregarded; 4% of the population are carriers; thus chances of carriers crossing =.04 x.04 =.0016 or 0.16%; (allow other ways of showing figures) probability of carrier cross producing a CF child is 1 in 4; thus expected incidence will be 0.16 = 0.04%; 4 (if say 100 = 0.04% allow 1 mark only, unless explained) max (iii) because new mutations (of the same type) are constantly happening; 1 (c) remove mucus by physiotherapy/thoracic massage/aspiration; diet control (to counteract pancreatic misfunction); gene therapy; max 2 Page 4 TOTAL 18

5 QUESTIONSHEET 8 ref to restitution nucleus; failure of chromosomes to separate during anaphase; most likely in mitosis in apical meristem; thus this part/sector of the plant would be tetraploid and would produce diploid gametes; since inbreeding these would produce tetraploid seed/offspring; which would breed on to produce more tetraploids; max 4 gametes of Coffea arabica will contain 22 chromosomes; gametes of other species/ancestral form will contain 11 chromosomes; thus accurate bivalent formation in meiosis cannot occur; so even if hybrid grows it will not be able to produce viable gametes; max 3 (b) genes from other stock will not be incorporated and so it becomes genetically stable; all plants will have basically the same genotype; good features/high yield/disease resistance/flavour/quick growth/any valid example, will be perpetuated/ not diluted by intrusion of other genes; polyploidy will be maintained; polyploids produce larger beans; max 4 TOTAL 11 QUESTIONSHEET 9 (a) pollen of the one species fertilises the ovule/embryo sac of the other species; seeds produced; these germinate/grow to produce offspring with one set of chromosomes from each of the parent species; ref to F 1 hybrid; ref to allopolyploidy; max 3 (b) no meiosis can occur; since (non-matching) chromosomes of orange and pummelo will not pair (in synapsis); thus haploid pollen/egg nuclei cannot be produced; seed develops purely by mitosis; max 2 (c) no variation due to meiosis/random assortment/chiasmata; no variation due to outbreeding/fertilisation; variation can only occur by mutation; max 2 (d) gene mutation/point mutation (of a gene); (reject polyploidy) ref to substitution/deletion/addition/inversion/translocation (of genes); 2 TOTAL 9 Page 5

6 QUESTIONSHEET 10 (a) change in the base sequence of the gene; due to substitution/deletion/addition/inversion/translocation of bases; alters the codon sequence of the gene; (thus) alters/may alter the amino acid sequence of the polypeptide made by the gene; max 3 sickle cell anaemia; haemophilia; (red-green) colour blindness; (allow other correct examples) max 2 non-disjunction; sets of chromosomes fail to segregate correctly to poles (in anaphase of meiosis); results in egg cell containing two of chromosome 21; when fertilised results in trisomy 21/Down s syndrome baby has three copies of chromosome 21; another cause of Down s syndrome is due to translocation of part of another chromosome onto the end of chromosome 21; max 3 TOTAL 8 QUESTIONSHEET 11 (iii) incidence remains very low/less than 2 per thousand/does not increase up the the age of 30 (years); steady increase/increases to 4 per thousand/incidence doubles between 30 and 35 (years); steeper increase/increases to 18 per thousand from 35 to 45 (years); 3 meiosis becomes less efficient as mother ages/random assortment/segregation to poles less efficient; older parents have undergone longer exposure to possible mutagens and so tend to have higher mutation rates; max 1 non-disjunction; failure of sets of chromosomes to segregate accurately (in anaphase of meiosis); for instance two of chromosome 21/18/13 may go to one pole and none to the other pole; (thus) egg may contain two copies of the chromosome; (thus) after fertilisation the zygote will have three copies of the chromosome; max 3 (b) ref amniocentesis; collect amniotic fluid (which contains fetal cells); centrifuge to collect fetal cells and then grow them in tissue culture; treat with colchicine/drug to hold chromosomes in metaphase/visible/spread out state; make smears and stain (with chromosome stain); photograph chromosomes of many nuclei, cut out and pair/ref karyotyping; max 3 TOTAL 10 QUESTIONSHEET 12 inversion; 1 A B 1 cross bandings show positions of stained/similar DNA/DNA bands marked by gene probes; cross bandings match very closely (in non-mutated part of chromosomes); even match in mutated region taking into account the inversion; since they have similar staining/marked DNA/similar DNA distribution they are probably closely related; max 3 after divergence from common stock/common ancestry; 1 Page 6 TOTAL 6

Human Chromosomes Section 14.1

Human Chromosomes Section 14.1 In Today s class. We will look at Human chromosome and karyotypes Autosomal and Sex chromosomes How human traits are transmitted How traits can be traced through entire families